- #36
Hans de Vries
Science Advisor
Gold Member
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JesseM said:http://grad.physics.sunysb.edu/~amarch/PHY5655.gif [Broken]
http://grad.physics.sunysb.edu/~amarch/ [Broken]
http://grad.physics.sunysb.edu/~amarch/Walborn.pdf [Broken]
Nice, experiment.
Note however that all results are just what one would expect from classical optics...
If you look at the pdf.
FIG 2: The interference pattern.
FIG 3: The interference pattern disappears after the insertion of the quarter-wave plates.
The pattern also disappears in classical optics: The Left plus Right hand polarized light
produces TWO interference patterns. One for horizontal and one for vertical polarized light.
The two patterns are 180 degrees displaced relative to each other and together they
sum up to the "non-interference" pattern of Figure 3.
FIG 4: The interference pattern reappears if a 45 degrees polarizer is inserted in the other beam.
This is actually one of the two interference patterns mentioned above. The polarizer
absorbes ~70% of the photons and 45 degrees polarized photons have the highest
change of getting through.
Therefor the -45 degrees photons at the dual split side have the highest change of
getting counted. What does happen now at the quarter-wave plates? Look at the table
in the middle of http://grad.physics.sunysb.edu/~amarch/ [Broken]
Code:
QWP1: 0 degrees --> Right polarized light, 90 degrees --> Left polarized light
QWP2: 90 degrees --> Left polarized light, 0 degrees --> Right polarized light
Under 45 (or -45) degrees they both produce linear polarized light. Either both horizontal
or both vertical light. Figure 4 therefore corresponds with one of the two interference
patterns adding up to the "non-interference" pattern of Figure 3.
FIG 5: A 180 degrees displaced pattern appears if photons which tend to have a -45 degrees polarization are counted.
This is the other of the two interference patterns adding up to Figure 3. The displacement
of the interference pattern is determined by classical optics. Figures 4 and 5 add up to
Figure 3 as they should do.
Regards, Hans
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